Determine the empirical and molecular formula of the unknown acid?

Combustion of 19.81 mg of an industrial acid produces 41.98 mg of CO2 and 6.45 mg of H2O. If 0.250 moles of the acid has a mass of 41.5 g, determine the empirical and molecular formula of the acid.

Combustion of 19.81 mg of an industrial acid produces 41.98 mg of CO2 and 6.45 mg of H2O. If 0.250 moles of the acid has a mass of 41.5 g, determine the empirical and molecular formula of the acid.

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Oct 18, 2017

Answer:

phtalic acid: #C_6H_4(COOH)_2#

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combustion of 19.81 mg of an industrial acid produces 41.98 mg of CO2 and 6.45 mg of H2O. If 0.250 moles of the acid has a mass of 41.5 g, determine the empirical and molecular formula of the acid.
The acid is an organic acid made of C, H and O with probably formula# C_nH_mO_2#

the acid has a molecular mass of #(41.5 g)/(0.25mol) =166g/(mol)# . dividing 19,81 mg for 166g/mol you obtain 0,119 mmol
If You divide 41.98 mg Di #CO_2# for 44g/mol that is its MM you obtain the mmol of #CO_2# svilupped from the copmbustion =0.954 mmol.

Making tha same procediment for water, you obtain:mmol water =#(6.45)/18=0.358# mmol of water
Now we divide the three numbers for the smaller one.

By dividing 0.954/0.119 you obtain 8 that is 8 mol of #CO_2# for 1 mol of acid, By dividing 0.358/0.119 You obtain 3 that is 3 mol of #H_2O# produced by the reaction that means 8 atoms of carbon (MA= 6X12= 96 and 6 of Hydrogen (MA= 6 xx 1 =6) in the initial organic acid. Oxygen will be ((166-96-6) =64 g that divided for 16 g/mol(MA O) givew 4 atoms of O in the molecula that will have the formula #C_8O_4H_6#.

It comes to my mind only the phtalic acid (two groups COOH attacked to one benzenic ring) with this formula: #C_6H_4(COOH)_2# .

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